Convergence of {fn} wrt to C(X) metric

[camillio]

Let C(X) be the set of continuous complex-valued bounded functions with domain X. Let {fn} be a sequence of functions on C. We say, that fn converges to f wrt to the metric of C(X) iff fn converges to f uniformly.

By definition of the uniform convergence, for any ε>0 there exists integer N s.t. n ≥ N implies |fn(x) - f(x)| ≤ ε for every x in X. Hence by the definition of supremum norm, the inequality is equivalent to sup(fn(x) - f(x)) ≤ ε.

I conclude, that:
1) the uniform convergence imposes that for every ε>0, there exists N from which on the functions fn and f are within this ε on whole X. Hence it always implies convergence wrt the metric of C(X), since exactly this ε can be used as the supremum of the distance fn and f.
2) What about the converse? I think, that if {fn} converges to f wrt to the metric of C(X), then the existence of such ε>0, being the supremum of the distance on whole X, implies the uniform convergence, as this ε can be used in its definition and the existence of N is guaranteed.

Am I right or not?

 

[Fredrik]

Is the question if it's true that $f_n\to f$ uniformly if and only if $f_n\to f$ with respect to the metric defined from the norm defined by $\|f\|=\sup_{x\in X}|f(x)|$ for all f in C(X)? If that's what you're asking, the answer is yes. This is how I would start a proof of that theorem:

Let $\varepsilon>0$ be arbitrary.

Suppose that $f_n\to f$ with respect to the norm (metric). Choose N such that for all n≥N,
$$\|f_n-f\|<\varepsilon.$$ This choice ensures that for all n≥N and all x in X,
$$|f_n(x)-f(x)| \dots $$
So $f_n\to f$ uniformly.

Can you fill in the details (at the dots)? The proof of the converse is very similar.

 

[camillio]

Thank you, Frederic. This was exactly what I was thinking about, and, roughly said, your answer follows the idea provided by me in (2). The goal was to ensure that I'm not missing something.